132_pdfsam_math 54 differential equation solutions odd - v...

Chapter 3With the chute open, the parachutist falls 1946.38 m. It takest∗seconds, wheret∗satisfiesx2(t∗) = 1946.38. Solving yields5g6t∗+5620−5g6(1−e−6t∗/5)= 1946.38⇒t∗≈236.884 (sec).Therefore, the parachutist will hit the ground aftert∗+t∗≈241.1 seconds.9.This problem is similar to Example 1 on page 110 of the text with the addition of a buoyancyforce of magnitude (1/40)mg. If we letx(t) be the distance below the water at timetandv(t) the velocity, then the total force acting on the object isF=mg−bv−140mg.We are givenm= 100 kg,g= 9.81 m/sec2, andb= 10 kg/sec. Applying Newton’s SecondLaw gives100dvdt= (100)(9.81)−10v−104(9.81)⇒dvdt= 9.56−(0.1)v .Solving this equation by separation of variables, we have

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